# Spectrum of a Ring

We have a functor Spec from Ring to Schemes:

(\text{Ring} \xrightarrow{\text{Spec}} \text{Schemes})

Schemes look like Fun(Ring, Set). So Spec sends a ring to a functor from Ring to Set.

(\text{Ring} \xrightarrow{\text{Spec}} (\text{Ring} \to \text{Set}))

Honestly, whenever I see “scheme” I replace it mentally with “algebraic curve,” but this is just because I don’t really get what a scheme is yet.

(R \xrightarrow{\text{Spec} (-)} \text{Spec(R)})

(R \xrightarrow{\text{Spec} (-)} (S \mapsto \text{hom}(R,S)))

(\text{Spec}) (R) (S) = (\text{Hom}_{\text{Ring}}(R, S))

Let say we have:

1. a ring (\mathbb{Z}[t]) (this is just the ring of polynomials with (t) as a variable and coefficients in (\mathbb{Z}))
2. an arbitrary ring (S).

What is (\text{hom}(\mathbb{Z}[t], S))?

Notice that we’re in Ring, so the members of (\text{hom}(\mathbb{Z}[t], S)) must be Ring homomorphisms. Let’s say we’re looking at a ring homomorphism (\phi: \mathbb{Z}[t] \to S).

I learned from Cris Moore that demanding examples is a useful practice, so: What is (\phi(7t^2-4t+3)) in simpler terms?

(\phi) is a ring homomorphism, so we know that (\phi(1) = 1_s). This implies that (\phi(n) = n_s).

(\phi(7t^2-4t+3) = \phi(7t^2) – \phi(4t) + \phi(3)) (= \phi(t)(\phi(7t) – \phi(4)) + \phi(3)) (= \phi(t) (\phi(t) 7_s – 4_s) + 3_s)

(\phi(t)) is not determined! We can pick it to be any element of S!

(\text{hom}(\mathbb{Z}[t], S) \simeq {\phi(t) \in S} = S)

In other words, (\text{Spec}(\mathbb{Z}[t])) is a forgetful functor from (S) as an object in (\text{Ring}) to its underlying set.

Exercise: show that (\mathbb{Z}[t]) is an initial object in an appropriate category, and that the trivial ring is the terminal object in the appropriate category.

#### But what is Spec?

It’s the spectrum of a ring! Intuitively, (\text{Spec} R) is the geometric object canonically associated to (R).

For example:

• (\text{Spec}(\mathbb{Z})) is “The Point”
• (\text{Spec}(\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))) is “The Circle”
• (\text{Spec}(\mathbb{Z}[x,x^{-1}]) is “Pairs of Invertible Elements” (e.g. a field with the origin removed)

Let’s look at (\text{Spec}(\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))(S)) when (S) is (\mathbb{R}), it’s obviously the circle BUT IT WORKS FOR (almost) ANY S. The concept of there being a canonical geometry associated to every ring is very exciting!

Sidenote:

When people say “elliptic curve” they might mean “a family of elliptic curves”, this is commonly written (C to \text{Spec} R) where (R) is the underlying coefficient ring.

For example, (y^2=4x^3+ax+b \mapsto a, b in R) is the family of elliptic curves of the form (y^2=4x^3+ax+b) over the coefficient ring (R).

I’m trying to figure out what happens when (S) is not something nice like (\mathbb{R}) or (\mathbb{C}). What is a circle in the coefficient ring of an elliptic curve?

Thanks to Aaron Mazel-Gee for walking me through the concept of (\text{Spec}) (all errors in this post are mine and not his).

Written on January 21, 2015